In a system A(s)⇌B(g)+3C(g), if the equilibrium concentration ofB increases by a factor of 8, it will cause the equilibrium concentration of C to change to

# In a system $\mathrm{A}\left(\mathrm{s}\right)⇌\mathrm{B}\left(\mathrm{g}\right)+3\mathrm{C}\left(\mathrm{g}\right)$, if the equilibrium concentration ofB increases by a factor of 8, it will cause the equilibrium concentration of C to change to

1. A

two times its original value

2. B

one half of its original value

3. C

eight times its original value

4. D

one-eight its original value

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### Solution:

$\left[\mathrm{B}{\right]}_{1}\left[\mathrm{C}{\right]}_{1}^{3}=\left[\mathrm{B}{\right]}_{2}\left[\mathrm{C}{\right]}_{2}^{3}$

It is given that $\left[\mathrm{B}{\right]}_{2}=8\left[\mathrm{B}{\right]}_{1}.$ Hence $\left[\mathrm{C}{\right]}_{2}=\frac{\left[\mathrm{C}{\right]}_{1}}{{8}^{1/3}}=\frac{\left[\mathrm{C}{\right]}_{1}}{2}$  